A radical ideal in a Noetherian ring is a finite intersection of minimal primes

Let $\mathcal{P}$ be the collection of radical ideals $J⊴R$ for which there exist finitely many prime ideals minimal over $J$ whose intersection equals $J$. By [L1], it is enough to fix a radical ideal $I$ and assume that every radical ideal strictly containing $I$ lies in $\mathcal{P}$.

If $I=R$, then no prime ideal contains $I$, and the empty intersection is $R$. So $I\in \mathcal{P}$ in this boundary case.

If $I$ is prime, then $I$ itself is the unique prime minimal over $I$, so $I\in \mathcal{P}$.

Assume now that $I$ is neither $R$ nor prime. Then $I$ is a proper radical ideal, so [L2] applies: choose $x,y\mathrm{\notin }I$ with $xy\in I$, and set $J=\sqrt{I+(x)}$ and $K=\sqrt{I+(y)}$. Because $I$ is radical and $x,y\mathrm{\notin }I$, both $J$ and $K$ strictly contain $I$. Hence the induction hypothesis places both $J$ and $K$ in $\mathcal{P}$, and [L2] also says that every prime minimal over $I$ is minimal over $I+(x)$ or over $I+(y)$.

Choose finite families of prime ideals minimal over $J$ and $K$ with $J={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r$ and $K={\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_s$. The product $(I+(x))(I+(y))$ is contained in $I$: expanding $(a+rx)(b+sy)$ gives terms in $I$ because $a,b\in I$ and $xy\in I$. Therefore $J\cap K=\sqrt{(I+(x))(I+(y))}\subseteq I$ by [L3]. The reverse inclusion $I\subseteq J\cap K$ is immediate, so $I={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r\cap {\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_s$.

By step 1.4, every prime minimal over $I$ belongs to the finite family $\mathcal{F}=\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r,{\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s\}$. Let $\mathcal{M}\subseteq \mathcal{F}$ be the subfamily of inclusion-minimal members. Every element of $\mathcal{M}$ is a prime ideal minimal over $I$, and every prime in $\mathcal{F}$ contains some member of $\mathcal{M}$ because $\mathcal{F}$ is finite. Removing a prime ideal that contains another one does not change an intersection, so the intersection over $\mathcal{M}$ is still $I$. Thus $I\in \mathcal{P}$.

Steps 1.2, 1.3, and 3.1 establish the induction step promised in step 1.1. Therefore every radical ideal of $R$ is an intersection of finitely many prime ideals minimal over it, and in particular has only finitely many minimal prime ideals.